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alex41 [277]
3 years ago
7

Given 2 points how would you write an algebraic representation

Mathematics
1 answer:
Ganezh [65]3 years ago
7 0

Answer:

Find the Equation of a Line Given That You Know Two Points it Passes Through. The equation of a line is typically written as y=mx+b where m is the slope and b is the y-intercept.

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WILL GIVE BRAINLIEST
Inessa [10]

Answer:

The answer is A.

Step-by-step explanation:

A sample space is the list of all possible outcomes. Only answer A displays all of them. In B, 2 wins/loss combinations are not possible if you play three games. The same goes for C and 4 win/loss combinations. While D has some possible outcomes, it does not list them all.

5 0
3 years ago
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A piece of steel pipe 12.5 feet long weighs 158 pounds. How much does a piece of the same steel pipe 18.75 feet long weigh?
jonny [76]

Answer:

18.75 feet of steel pipe weighs 237 pounds  

explanation:

find out how much 1 feet weighs than u can find out how much 18.75 feet.

by doing it like this  

158÷ 12.5

=12.64

1 feet equals 12.64 pounds

12.64× 18.75

=237 pounds this is the answer

i do apologize if it wrong

8 0
4 years ago
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G(x)=3x^2-5x-9 find g(x+2)
Anna35 [415]
Answer: 3x^2 + 7x -15
I dunno if u want me to factor it tho
7 0
3 years ago
I've wasted 42 points already. Can someone PLZ just answer my Question and stop wasting my points. PLz, this is due. I need help
Misha Larkins [42]

Step-by-step explanation:

You can put the value of x into the equation to solve for y. Hopes this works.

5 0
3 years ago
(cotx+cscx)/(sinx+tanx)
Butoxors [25]

Answer:   \bold{\dfrac{cot(x)}{sin(x)}}

<u>Step-by-step explanation:</u>

Convert everything to "sin" and "cos" and then cancel out the common factors.

\dfrac{cot(x)+csc(x)}{sin(x)+tan(x)}\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)}{1}+\dfrac{sin(x)}{cos(x)}\bigg)\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg[\dfrac{sin(x)}{1}\bigg(\dfrac{cos(x)}{cos(x)}\bigg)+\dfrac{sin(x)}{cos(x)}\bigg]\\\\\\\bigg(\dfrac{cos(x)}{sin(x)}+\dfrac{1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)cos(x)}{cos(x)}+\dfrac{sin(x)}{cos(x)}\bigg)

\text{Simplify:}\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)cos(x)+sin(x)}{cos(x)}\bigg)\\\\\\\text{Multiply by the reciprocal (fraction rules)}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)cos(x)+sin(x)}\bigg)\\\\\\\text{Factor out the common term on the right side denominator}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)(cos(x)+1)}\bigg)

\text{Cross out the common factor of (cos(x) + 1) from the top and bottom}:\\\\\bigg(\dfrac{1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)}\bigg)\\\\\\\bigg(\dfrac{1}{sin(x)}\bigg)\times cot(x)}\qquad \rightarrow \qquad \dfrac{cot(x)}{sin(x)}

6 0
4 years ago
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